about the metastable state processing by 2 stage FF method

Hi ,
As we all know ,inorder to avoid the metastable state transfer, usually we will let the A clk domain's signal pass through the two FF which are trigerred by the B clk.
since the two FF use same clk, so the second FF will get an stable signal from first FF .
my question is when the first FF get an metastable signal from A clk domain ,
what's happened on it's output ?
I guess maybe there's short term metastable signal on it's output and then change to a stable signal , but can't make sure it's 1 or 0 .
is that so as I suppose ?
thanks .

Hi,
Yes. You are right. First flop may go in metastable condition. In this case, output will be unpredictable, it might be oscillating between some voltage levels, or it might be moving to some voltage level( i.e. either Vcc or Vss).
While one use the synchronizer, one has to make sure that the first flip flop will resolved to some value(it can resolve to either 0 or 1) before the clk edge to second flop. i.e. for the first flop it has 1 clk period to get resolved. If this doesn't happen and metastable state of FF1, persist even after clk period time, second FF will either sample it inappropriately or second FF may also go in metastable. But in practice the chance of going second FF in metastable is very very less.(depends on MTBF of synchronizer).

One should make design by keeping in mind that synchronizer is going to take 2 clk cycles to transfer the value of other clk domain faithfully.
I hope this will help you.

Thanks viju, u help me make it clearer.
so that's why the 2 ff type synchronizer is called 'decrease the metastable probability ' ,
because the second ff also have the probability to get the metastable signal from the 1st ff , just it's probability is very small . and depends on the electronic performance of 1st ff .
I got it , thank you again .